Hirsutism, occurs due to abnormally high androgen levels
Step-by-step explanation:
I hope it's correct...
Answer:
BC = 22
Step-by-step explanation:
BD = 67
Bc = 3x - 2
CD = 4x + 13
BC + CD = BD (Given)
3x - 2 + 4x + 13 = 67
3x + 4x -2 + 13 = 67
7x + 11 = 67
7x = 67 - 11
7x = 56
x = 56 ÷ 7
x = 8
now instead of x put 8
Bc = 3x - 2
BC = 3(8) -2
BC = 24 -2
BC = 22
Answer: 2.5 ft.
Step-by-step explanation: We know that the wall is 10ft and each poster is 1 1/4 ft, so subtract both posters from the wall, 10-2.5 = 7.5, then divide 7.5 by 3 since we need 3 lengths, and we get 2.5ft. So for x, it would be 2.5ft.
x=2.5
Hope this helps!
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
